# Show that every local homeomorphism is continuous and open therefore bijective local homeomorphism is a homeomorphism

Follow up on another question I asked recently: Topology: Show restriction of continuous function is continuous, and restriction of a homeomorphism is a homeomorphism

Definition: Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces. A function ${\displaystyle f:X\to Y\,}$ is a local homeomorphism if for every point $x \in X$ there exists an open set $U \subseteq X$ containing $x$ and an open set $V \subseteq Y$ such that the restriction ${\displaystyle f|_{U}:U\to V\,}$ is a homeomorphism.

This definition is a bit alarming because it starts with "...if for every point $x \in X$ there exists an open set $U \in \mathcal{T}$...", makes it seem like a property of the underlying space. Can we always find an open $U$? But anyways.

Objective: Show that every local homeomorphism is continuous and open therefore bijective local homeomorphism is a homeomorphism

Proof: (Honestly not sure what I am doing but proceed regardless)

Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces and function ${\displaystyle f:X\to Y\,}$ is a local homeomorphism. We will show that $f$ is continuous and open.

First show $f$ is continuous.

$f$ is continuous if for all $V \in \mathcal{J}, f^{-1}(V) \in \mathcal{T}$. Take some $V \in \mathcal{J}$, then $V$ is a subspace equipped with subspace topology $\mathcal{J}_V = \{V \cap W| W \in \mathcal{J}\}$.